- True or false: A 100 kilogram moon rock has more inertia on Earth than it does on the Moon.

Un cuerpo de m=0,5 Kg se desplaza horizontalmente con v=4m/s y luego de un lapso de tiempo se mueve con v=20 m/s. ¿cual ha sido

Lilit [14]

Responder:

<h2>64 Julios </h2>

Explicación:

La energía cinética se expresa mediante la fórmula KE = 1 / 2mv² donde;

m es la masa del cuerpo

v es la velocidad del objeto

Dado que el cuerpo se mueve horizontalmente con v = 4 m / sy después de un período de tiempo se mueve con v = 20 m / s, entonces la variación en la velocidad será de 20 m / s - 4 m / s = 16 m / s.

Parámetros dados

masa del objeto m = 0,5 kg

Variación de velocidad = 16 m / s

Variación de la energía cinética = 1/2 * 0,5 * 16²

Variación de la energía cinética = 1/2 * 0,5 * 256

Variación de la energía cinética = 0,5 * 128

<em>Variación de la energía cinética = 64 Julios</em>

7 0

3 years ago

A train travels north at a speed of 50 m/s.

Gala2k [10]

The apparent velocity is B) 48 m/s north

Explanation:

Here we have a problem of relativity of velocities.

In fact, the train is travelling north at a speed of

$v_t = 50 m/s$

where this velocity is measured with respect to the ground.

At the same time, a passenger on the train is walking towards the rear (so, south) at a velocity of

$v'=2 m/s$

where this velocity is measured with respect to the train, which is in motion in the opposite direction.

Therefore, the apparent velocity of the passenger with respect to an observer standing on the ground is:

$v=V_t - v' = 50 - 2 = 48 m/s$

And the direction is north, since this number is positive.

Learn more about velocity:

brainly.com/question/5248528

#LearnwithBrainly

8 0

3 years ago

At the moment t = 0, a 20.0 V battery is connected to a 5.00 mH coil and a 6.00 Ω resistor. (a) Immediately thereafter, how does

insens350 [35]

(a) On the coil: 20 V, on the resistor: 0 V

The sum of the potential difference across the coil and the potential difference across the resistor is equal to the voltage provided by the battery, V = 20 V:

$V = V_R + V_L$

The potential difference across the inductance is given by

$V_L(t) = V e^{-\frac{t}{\tau}}$ (1)

where

$\tau = \frac{L}{R}=\frac{0.005 H}{6.00 \Omega}=8.33\cdot 10^{-4} s$ is the time constant of the circuit

At time t=0,

$V_L(0) = V e^0 = V = 20 V$

So, all the potential difference is across the coil, therefore the potential difference across the resistor will be zero:

$V_R = V-V_L = 20 V-20 V=0$

(b) On the coil: 0 V, on the resistor: 20 V

Here we are analyzing the situation several seconds later, which means that we are analyzing the situation for

$t >> \tau$

Since $\tau$ is at the order of less than milliseconds.

Using eq.(1), we see that for $t >> \tau$ , the exponential becomes zero, and therefore the potential difference across the coil is zero:

$V_L = 0$

Therefore, the potential difference across the resistor will be

$V_R = V-V_L = 20 V- 0 = 20 V$

(c) Yes

The two voltages will be equal when:

$V_L = V_R$ (2)

Reminding also that the sum of the two voltages must be equal to the voltage of the battery:

$V=V_L +V_R$

And rewriting this equation,

$V_R = V-V_L$

Substituting into (2) we find

$V_L = V-V_L\\2V_L = V\\V_L=\frac{V}{2}=10 V$

So, the two voltages will be equal when they are both equal to 10 V.

(d) at $t=5.77\cdot 10^{-4}s$

We said that the two voltages will be equal when

$V_L=\frac{V}{2}$

Using eq.(1), and this last equation, this means

$V e^{-\frac{t}{\tau}} = \frac{V}{2}$

And solving the equation for t, we find the time t at which the two voltages are equal:

$e^{-\frac{t}{\tau}}=\frac{1}{2}\\-\frac{t}{\tau}=ln(1/2)\\t=-\tau ln(0.5)=-(8.33\cdot 10^{-4} s)ln(0.5)=5.77\cdot 10^{-4}s$

(e-a) -19.2 V on the coil, 19.2 V on the resistor

Here we have that the current in the circuit is

$I_0 = 3.20 A$

The problem says this current is stable: this means that we are in a situation in which $t>>\tau$ , so the coil has no longer influence on the circuit, which is operating as it is a normal circuit with only one resistor. Therefore, we can find the potential difference across the resistor using Ohm's law

$V=I_0 R = (3.20 A)(6.0 \Omega)=19.2 V$

Then the battery is removed from the circuit: this means that the coil will discharge through the resistor.

The voltage on the coil is given by

$V_L(t) = -V e^{-\frac{t}{\tau}}$ (1)

which means that it is maximum at the moment when the battery is disconnected, when t=0:

$V_L(0)=.V$

And V this time is the voltage across the resistor, 19.2 V (because the coil is now connected to the resistor, not to the battery). So, the voltage across the coil will be -19.2 V, and the voltage across the resistor will be the same in magnitude, 19.2 V (since the coil and the resistor are connected to the same points in the circuit): however, the signs of the potential difference will be opposite.

(e-b) 0 V on both

After several seconds,

$t>>\tau$

If we use this approximation into the formula

$V_L(t) = -V e^{-\frac{t}{\tau}}$ (1)

We find that

$V_L = 0$

And since now the resistor is directly connected to the coil, the voltage in the resistor will be the same as the coil, so 0 V. This means that the coil has completely discharged, and current is no longer flowing through the circuit.

7 0

3 years ago

how does the frequency of a radio wave compare to the frequency of the vibrating electrons that produce it?

HACTEHA [7]

Answer:

mass

Explanation:

3 0

3 years ago

In what way are gravitational and electrical forces similar?

Nadusha1986 [10]

Answer:

D. Both occur between objects independently whether they are in contact or not.

Explanation:

- The gravitational force is a force that is exerted between two (or more) objects having mass. This force is always attractive and its magnitude is given by

$F=G\frac{m_1 m_2}{r^2}$

where G is the gravitational constant, m1 and m2 are the two masses, and r is the distance between the two masses.

- The electrical force is a force that is exerted between two (or more) objects having electrical charge. It can be either attractive or repulsive, depending on the sign of the two charges, and its magnitude is given by

$F=k\frac{q_1 q_2}{r^2}$

where k is the Coulomb's constant, q1 and q2 are the two charges, and r the distance between the two charges.

Looking at both formulas, we see that the two forces are present even when the two objects are not in contact with each other (in fact, r can assume any value in the formula). They are said to be non-contact forces. Therefore, the correct option is

D. Both occur between objects independently whether they are in contact or not.

6 0

3 years ago